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Mechanical Properties of Fluids — Physics STD 11 Science — Question

Gujarat BoardEnglish MediumSTD 11 SciencePhysicsMechanical Properties of Fluids5 Marks
Question
Water stands at a depth H in a tank, whose side walls are vertical. A hole is made on one of the walls at a depth h below the water surface. Find at what distance from the foot of the wall does the emerging stream of water strike the floor and for what value of h this range is maximum.
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Answer


The situation is shown in fig. Here, we have,$\upsilon_\text{A}=\sqrt{2\text{gh}}\dots\text{(i)}$
$(\text{H}-\text{h})=\frac12\text{gt}^2\dots\text{(ii)} $
The distance R is given by$\text{R}=\upsilon_\text{A}\times\text{t}\dots\text{(iii)}$
From equation (ii) $\text{t}=\sqrt{\Big(\frac{2(\text{H}-\text{h})}{\text{g}}\Big)}\dots\text{(iv)}$ Substituting the value of $v_A$ from equation (1) and the value of t from equation (4) in equation (3), we get,$\text{R}=\sqrt{(2\text{gh})}\times\sqrt{\Big\{\frac{2(\text{H}-\text{h})}{\text{g}}\Big\}}$
$=\sqrt{\{\text{h}(\text{H}-\text{h}\}}.$
The range R will be maximum when$\frac{\text{dR}}{\text{dh}}=0$
$\therefore2\cdot\frac12\text{h}^{-\frac12}(\text{H}-\text{h})^{\frac12}\cdot\frac12(\text{H}-\text{h})^{-\frac12}=0$
Solving we get $\text{h}=\frac{\text{H}}2$

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